wave mechanics lecture

8/12/2019 Wave Mechanics Lecture

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PHYS201

Wave Mechanics

Semester 1 2009

J D Cresser

[email protected] C5C357

Ext 8913

Semester 1 2009 PHYS201 Wave Mechanics 1 / 86
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Quantum Mechanics and Wave Mechanics I

Quantum mechanics arose out of the need to provide explanations for a range of

physical phenomena that could not be accounted for by classical physics:

Black body spectrum

Spectra of the elements

photoelectric effect

Specific heat of solids . . .

Through the work of Planck, Einstein and others, the idea arose that electromagnetic

and other forms of energy could be exchanged only in definite quantities (quanta)

And through the work of De Broglie the idea arose that matter could exhibit wave like

properties.

It was Einstein who proposed that waves (light) could behave like particles (photons).

Heisenberg proposed the first successfulquantum theory, but in the terms of the

mathematics of matrices matrix mexhanics.Semester 1 2009 PHYS201 Wave Mechanics 2 / 86
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Quantum Mechanics and Wave Mechanics II

Schrodinger came up with an equation for the waves predicted by de Broglie, and that

was the start of wave mechanics.

Schrodinger also showed that his work and that of Heisenbergs were mathematicallyequivalent.

But it was Heisenberg, and Born, who first realized that quantum mechanics was a

theory of probabilities.

Something that Einstein would never accept, even though he helped discover it!

Through the work of Dirac, von Neumann, Jordan and others, it was eventually shown

that matrix mechanics and wave mechanics were but two forms of a more fundamental

theory quantum mechanics.

Quantum mechanics is a theory of information

It is a set of laws about the information that can be gained about the physical world.

We will be concerned with wave mechanics here, the oldest form of quantum

mechanics.Semester 1 2009 PHYS201 Wave Mechanics 3 / 86
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The Black Body Spectrum I

Environment at temperature T

Black body at temperature T

f

S

Observed spectrum

Rayleigh-Jeans spectrum

The first inkling of a new physics lay in

understanding the origins of the black

body spectrum:

A black body is any object that absorbsall radiation, whatever frequency, that fallson it.

The blackbody spectrum is the spectrum

of the radiation emitted by the object when

it is in thermal equilibrium with its

surroundings.

Classical physics predicted the

Rayleigh-Jeans formula which diverged at

high frequencies the ultra-violet

catastrophe.

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The Black Body Spectrum II

Planck assumed that matter could absorb or emit black body radiation energy of

frequencyfin multiples of hf whereh was a parameter to be taken to zero at the end

of the calculation.

But he got the correct answer for a small but non-zero value ofh:

S(f, T) = 8hf3

c31

exp(hf/kT) 1 .

where

h =6.6218 1034Joules-secnow known as Plancks constant.

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Einsteins hypothesis

Einstein preferred to believe that the formula hfwas a property of the electromagnetic

field

Planck thought it was a property of the atoms in the blackbody.

Einstein proposed that light of frequencyfis absorbed or emitted in packets (i.e.

quanta) of energy Ewhere

E =hf Einstein/Planck formula

Sometime later he extended this idea to say that light also had momentum, and that it wasin fact made up of particles now called photons.

Einstein used this idea to explain the photo-electric effect (in 1905).

A first example of wave-particle duality: light, a form of wave motion, having

particle-like properties.

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The Bohr model of the hydrogen atom I

n=

n=2

n=

n=

emitted

photons

Rutherford proposed that an atom consisted of a

small positively charged nucleus with electrons in

orbit about this nucleus.

According to classical EM theory, the orbiting

electrons are accelerating and therefore ought to

emit EM radiation.

The electron in a hydrogen atom should spiral into thenucleus in about1012 sec.

Bohr proposed existence of stable orbits (stationary states) of radius rsuch that

angular momentum =pr =n h2

, n =1, 2, 3, . . .

Leads to orbits of radii rn and energiesEn:

rn =n2 0h

2

e2meEn =

1

n2mee

4

82

0

h2 n =1, 2, . . .

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The Bohr model of the hydrogen atom II

Bohr radius radius of lowest energy orbit a0 =r1 =0h

2

e2me=0.05 nm.

The atom emits EM radiation by making a transition between stationary states,

emitting a photon of energy hf where

hf =EmEn = mee

4

820

h2

1

n2 1

m2

Einstein later showed that the the photon had to carry both energy andmomentum henceit behaves very much like a particle.

Einstein also showed that the time at which a transition will occur and the direction ofemission of the photon was totally unpredictable.

The first hint of uncaused randomness in atomic physics.

Bohrs model worked only for hydrogen-like atoms

It failed miserably for helium

It could not predict the dynamics of physical systems

But it broke the ice, and a search for classical physics answerswasabandoned.Semester 1 2009 PHYS201 Wave Mechanics 8 / 86
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De Broglies Hypothesis

In 1924 Prince Louis de Broglie asked a philosophical question:

If waves can sometimes behave like particles, can particles sometimes exhibitwave-like behaviour?

To specify such a wave, need to know its

frequencyf use Einsteins formulaE =hf

wavelength make use of another idea of Einsteins

Special relativity tells us that the energyEof a zero rest mass particle (such as a

photon) is given in terms of its momentum p by

E =pc =hf.

Using the fact thatf =cgives pc =hc/

which can be solved for =h/p.

As there isno mention in this equation of anything to do with light, de Broglie assumed

it would hold for particles of matter as well.Semester 1 2009 PHYS201 Wave Mechanics 9 / 86
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Evidence for wave properties of matter

50

interference maximum

of reflected electrons

nickel crystal

nc ent electron

beam of energy

54 eV

=2r3

3

r3

In 1926 Davisson and Germer accidently showed

that electrons can be diffracted (Bragg diffraction)

Fired beam of electrons at nickel crystal andobserved the scattered electrons.

They observed a diffraction pattern identical to thatobserved when waves (X-rays) were fired at the

crystal.

The wavelengths of the waves producing thediffraction pattern was identical to that predicted bythe de Broglie relation.

G P Thompson independently carried out thesame experiment. His father (J J Thompson) hadshown that electrons were particles, the sonshowed that they were waves.

Fitting de Broglie waves around a circle gives Bohrs quantization condition

n =2rn =

prn =n

h

2Semester 1 2009 PHYS201 Wave Mechanics 10 / 86
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The Wave Function

Accepting that these waves exist, we can try to learn what they might mean.

We know the frequency and the wavelength of the wave associated with a particle of

energyEandp.

Can write down various formulae for waves of givenf and.

Usually done in terms of angular frequency and wave number k:

=2f k =2/

In terms of =h/2

E =hf =(h/2)(2f) and p =h/ =(h/2)(2/)

i.e. E = and p = k

Some possibilities for these waves:

(x, t) =A cos(kx t)or =A sin(kx t)or =Aei(kxt)

or = . . .

But what is doing the waving?? What are these waves madeof ?Semester 1 2009 PHYS201 Wave Mechanics 11 / 86
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What can we learn from the wave function? I

To help us understand what might be, let us reverse the argument

I.e. if we are given (x, t) =A cos(kx t), what can we learn about the particle?

This wave will be moving with aphase velocitygiven byvphase =

k.

Phase velocity is the speed of the crests of the wave.

UsingE = = 12

mv2 andp = k =mv wherev is the velocity of the particle we get:

vphase =

k=

E

p=

12

mv2

mv=

12

v

That is, the wave is moving with half the speed of its associatedparticle.Semester 1 2009 PHYS201 Wave Mechanics 12 / 86
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What can we learn from the wave function? II

Thus, from the phase velocity of the wave, we can learn the particles velocity via

v =2vphase.

What else can we learn Eandp, of course.

But what about the position of the particle?

(x, t)

x

vphase

This is a wave function of constant amplitude and wavelength. The wave is the sameeverywhere and so there is no distinguishing feature that could indicate one possible

position of the particle from any other.

Thus,we cannot learnwherethe particle is from this wave function.

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Wave Packets I

We can get away from a constant amplitude wave by adding together many waves with

a range of frequencies and wave numbers.

cos(5x) + cos(5.25x)

cos(4.75x) + cos(4.875x) + cos(5x) + cos(5.125x) + cos(5.25x)

cos(4.

8125x) + cos(4.

875x) + cos(4.

9375x) + cos(5x) + cos(5.

0625x) + cos(5.

125x) + cos(5.

1875x)

An integral over a continuous range of wave numbers produces a single wave packet.


W P k t II
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Wave Packets II

To form a single wave packet we need to carry out the sum (actually, an integral):

(x, t) =A(k1) cos(k1x

1t) +A(k2) cos(k2x

2t) + . . .

=

n

A(kn) cos(knx nt)

1'-1"

8"

(a) (b)

2k

2x

xkk

(x, t)A(k)

(a) The distribution of wave numbers kof harmonic waves contributing to the wave

function (x, t). This distribution is peaked about kwith a width of2k.

(b) The wave packet (x, t)of width2xresulting from the addition of the waves with

distributionA(k). The oscillatory part of the wave packet (the carrier wave) has wave

numberk.Semester 1 2009 PHYS201 Wave Mechanics 15 / 86

W k t III
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Wave packets III

2xvgroup

x

(x, t)

Particle in here

somewhere??

The wave function of a wave

packet is effectively zero

everywhere except in a region ofsize2x.

Reasonable to expect particle to

be found in region where wave

function is largest in magnitude.

The wave packet ought to behave in some way like its associated particle e.g. does it

move like a free particle.

We can check the speed with which the packet moves by calculating itsgroupvelocity

vgroup

= ddkk=k.From E =

p2

2mwe get =

2k2

2m

and hence =k2

2m a dispersion relation which givesvgroup =

k

m=

p

m

The wave packet moves as a particle of mass m and momentump = k.Semester 1 2009 PHYS201 Wave Mechanics 16 / 86

Heisenberg Uncertaint Relation I
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Heisenberg Uncertainty Relation I

Thus the wave packet is particle-like:

It is confined to a localized region is space . . .

It moves like a free particle.

But at the cost of combining together waves with wave numbers over the range

k kkk+ k

This corresponds to particle momenta over the rangep p p p + p

and which produces a wave packet that is confined to a region of size 2x.

I.e. the wave packet represents a particle to a certain extent, but it does not fix its

momentum to better than

pnor fix its position to better than

x.

The bandwidth theorem (or, Fourier theory) tells us that xk1.

Multiplying by gives theHeisenberg Uncertainty Relation.

xp


Heisenberg Uncertainty Relation II
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Heisenberg Uncertainty Relation II

Heisenbergs uncertainty relation is a direct result of the basic principles of quantum

mechanics

It is saying something very general about the properties of particles

No mention is made of the kind of particle, what it is doing, how it is created . . .

It is tells us that we can have knowledge of the position of a particle to within an uncertaintyofx, and its momentum to within an uncertaintyp, with xp

If we know the position exactly x =0, then the momentum uncertainty is infinite and vice versa

Itdoes notsay that the particle definitely has a precise position and momentum, but wecannot measure these precise values.

It can be used to estimate size of an atom, the lowest energy of simple harmonic

oscillator, temperature of a black hole, the pressure in the centre of a collapsing star . .

.


The size of an atom
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The size of an atom

p

a

e ectron

pro on

Can estimate the size of an atom by use of the

uncertainty relation.

The position of the electron with respect to the

nucleus will have an uncertainty x =a

The momentum can swing betweenp andpso willhave an uncertainty of p =p

Using the uncertainty relation in the form xp we getap .

The total energy of the atom is then

E = p2

2m e

2

40a

2

2ma2 e

2

40a

The radiusa that minimizes this is found from dE

da=0, which gives

a 402

me2 0.05nm (Bohr radius) and Emin 12

me4

(40)22 13.6eV.


THE TWO SLIT EXPERIMENT
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THE TWO SLIT EXPERIMENT

Shall be considered in three forms:

With macroscopic particles (golf balls);

With classical waves (light waves);

With electrons.

The first two merely show us what we would expect in our classical world.

The third gives counterintuitive results with both wave and particle characteristics that

have no classical explanation.


An Experiment with Golf balls: Both Slits Open
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An Experiment with Golf balls: Both Slits Open.

We notice three things about this experiment:

Erratic

golfer

Slit 1

Slit 2

Fence

The golf balls arrive in lumps: for each golfball hit that gets through the slits, there is a

single impact on the fence.

The golf balls arrive at random.

If we wait long enough, we find that the golf ball arrivals tend to form a pattern:

Erratic

golfer

Slit 1

Slit 2

Fence

This is more or less as we might expect:

The golf balls passing through hole 1 pileup opposite hole 1 . . .

The golf balls passing through hole 2 pileup opposite hole 2.


An Experiment with golf balls: one slit open
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An Experiment with golf balls: one slit open

.

If we were to perform the experiment with one or the other of the slits closed, would

expect something like:

Slit 1

blocked

Slit 2

blocked

i.e. golf ball arrivals accumulate opposite

the open slits.

expect that the result observed with both

slits open is the sum of the results

obtained with one and then the other slit

open.

Express this sum in terms of the probabilit

of a ball striking the fence at position x.


Probability for golf ball strikes
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Probability for golf ball strikes

Can sketch in curves to represent the probability of a golf ball striking the back fence at

positionx.

Thus, for instance,

P12(x)x =probability of a golf ball striking screen between x and x + x when both slits

are open.

P1(x) P2(x) P12(x)

Now make the claim that, for golf balls: P12(x) =P1(x) + P2(x)


Probability distributions of golf ball strikes . . .
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Probability distributions of golf ball strikes . . .Golf balls behave like classical particles

The fact that we add the probabilities P1(x)and P2(x)to getP12(x) is simply stating:

The probability of a golf ball that goes through slit 1 landing in(x,x + x)is completelyindependent of whether or not slit 2 is open.

The probability of a golf ball that goes through slit 2 landing in(x,x + x)is completelyindependent of whether or not slit 1 is open.

The above conclusion is perfectly consistent with our classical intuition.

Golf balls are particles: they arrive in lumps;

They independently pass througheitherslit 1orslit 2 before striking the screen;

Their random arrivals are due to erratic behaviour of the source (and maybe randombouncing around as they pass through the slit), all of which, in principle can be measuredand allowed for and/or controlled.


An Experiment with Waves
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An Experiment with Waves

Now repeat the above series of experiments with waves shall assume light waves of

wavelength.

Shall measure the time averaged intensity of the electric field E(x, t) =E(x) cos tof

the light waves arriving at the pointx on the observation screen.

The time averaged intensity will then be

I(x) =E(x, t)2 = 12E(x)2

Can also work with acomplexfield

E(x, t) = 12

E(x)eit

so that

E(x, t) =

2ReE(x, t)and

I(x) =E(x, t)E(x, t) =|E(x, t)|2


An Experiment with Waves: One Slit Open
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p p

I1(x)I2(x)

I1(x)and I2(x)are the intensities of the waves passing through slits 1 and 2

respectively and reaching the screen atx. (They are just the central peak of a single

slit diffraction pattern.)

Waves arrive on screen all at once, i.e. they do not arrive in lumps, but . . .

Results similar to single slits with golf balls in that the intensity is peaked directly

opposite each open slit.


An Experiment with Waves: Both Slits Open
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p p

I12(x)

Two slit interference

pattern

Interference if both slits are open:

I12(x) =(E1(x, t) +E2(x, t))2 =I1(x) +I2(x) + 2E1E2cos2dsin

=I1(x) +I2(x) + 2

I1(x)I2(x) cos

Last term is theinterferenceterm which explicitly depends on slits separationd the

waves probe the presence of both slits.

We notice three things about these experiments:

The waves arrive everywhere at once, i.e. they do not arrive in lumps.

The single slit result for waves similar to the single slit result for golf balls.

We see interference effects if both slits open for waves, but notforgolf balls.


An experiment with electrons
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p

Now repeat experiment once again, this time with electrons.

Shall assume a beam of electrons, all with same energy Eand momentump incident

on a screen with two slits.

Shall also assume weak source: electrons pass through the apparatus one at a time.

Electrons strike a fluorescent screen (causing a flash of light) whose position can be

monitored.


An experiment with electrons: one slit open


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Electron gun

Slit 1

Slit 2

fluorescent flashes

P2(x)

P1(x)

With one slit open observe electrons striking fluorescent screen in a random fashion,

but mostly directly opposite the open slit exactly as observed with golf balls.

Can construct probability distributionsP1(x)and P2(x)for where electrons strike, as for

golf balls.

Apparent randomness in arrival of electrons at the screen could be put down to

variations in the electron gun (cf. erratic golfer for golf balls).


An experiment with electrons: both slits open
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P12(x) Two slit interfer-ence pattern!Electron gun

The following things can be noted:

Electrons strike the screen causing individual flashes, i.e. they arrive as particles, just asgolf balls do;

They strike the screen at random same as for golf balls.

Can construct probability histogramsP1(x),P2(x)and P12(x)exactly as for golf balls.

Find that the electron arrivals, and hence the probabilities, form an interferencepattern

as observed with waves:

P12(x) =P1(x) + P2(x) + 2

P1(x)P2(x) cos2dsin

P1(x) + P2(x) The expected result for particles.


An experiment with electrons: both slits openI t f f d B li


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Interference of de Broglie waves

Problem: we have particles arriving in lumps, just like golf balls, i.e. one at a time at

localised points in space . . .

but the pattern formed is that of waves . . .

and waves must pass through both slits simultaneously, and arrive everywhere on the

observation screen at once to form an interference pattern.

Moreover, from the pattern can determine that =h/p the de Broglie relation.

Interference of de Broglie waves seems to have occurred here!


What is going on here?


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If electrons are particles (like golf balls) then each one must go through either slit 1 or

slit 2.

A particle has no extension in space, so if it passes through slit 1 say, it cannot possibly beaffected by whether or not slit 2, a distance daway, is open.

Thats why we claim we ought to find that

P12(x) =P1(x) + P2(x)

but we dont.

In fact, the detailed structure of the interference pattern depends on d, the separation

of the slits.

So, if an electron passes through slit 1, it must somehow become aware of the presence ofslit 2 a distancedaway, in order to know where to land on the observation screen so as to

produce a pattern that depends on d.

i.e. the electrons would have to do some strange things in order to know about thepresence of both slits such as travel from slit 1 to slit 2 then to the observation screen.

Maybe we have to conclude that it is not truethat the electrons go through either slit 1

or slit 2.


Watching the electrons


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We can check which slit the electrons go through by watching next to each slit and

taking note of when an electron goes through each slit.

If we do that, we get the alarming result that the interference pattern disappears we

regain the result for bullets.

It is possible to provide an explanation of this result in terms of the observation

process unavoidably disturbing the state of the electron.

Such explanations typically rely on invoking the Heisenberg Uncertainty Principle.


The Heisenberg Microscope


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electron momentum pde Broglie wavelength

= h/p

Weak incident light field: minimum of one

photon scatters off electron.

In order to distinguish the images of the

slits in the photographic plate of the

microscope, require wavelength of photon

at leastl

d. Momentum of photon is

then h/d

Photon-electron collision gives electron a

sideways momentum kick of p =h/d.

So electron deflected by angle of

p/p(h/. d/h) =d/

is approximately the angular separation between an interference maximum and a

neighbouring minimum!


An uncertainty relation explanation
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Recall the Heisenberg uncertainty relation for a particle:

xp 12

where xis the uncertainty in position and pis the uncertainty in momentum.

We could also argue that we are trying to measure the position of the electron to better

than an uncertainty

xd, the slit separationThis implies an uncertainty in sideways momentum of p/2d

This is similar to (but not quite the same as) the deflection p =h/dobtained in themicroscope experiment a more refined value for xis needed.

I.e., the requirement that the uncertainty principle (a fundamental result of wave mechanics)has to be obeyed implies that the interference fringes are washed out.

What if we observed the position of the electron by some other means?


The uncertainty relation protects quantum mechanics
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Foranymeans of measuring which slit the electron passes through.

The uncertainty relation gets in the way somewhere.

We always find that

Any attempt to measure which slit the electron passes through results in no

interference pattern!

The uncertainty relation appears to be a fundamental law that applies to allphysical

situations to guarantee that if we observe which path, then we find no interference.

But: it also appears that a physical intervention, an observation is needed to

physically scrambled the electrons momentum.

In fact, this is not the case!!

There existinteraction-free measurementsthat tell which path without touching theelectrons!


Interference of de Broglie waves I
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Recall: the probability distribution of the electron arrivals on the observation screen is

P12(x) =P1(x) + P2(x) + 2

P1(x)P2(x) cos

This has the same mathematical form as interference of waves:

I12(x) =(E1(x, t) +E2(x, t))2 =I1(x) +I2(x) + 2E1E2cos 2dsin

=I1(x) +I2(x) + 2

I1(x)I2(x) cos

Note that the interference term occurs because waves from two slits are added, and thenthe square is taken of the sum.

The time average is not an important part of the argument it is there simply as we aredealing with rapidly oscillating light waves.

So, it looks like the interference pattern for electrons comes about because waves from

each slit are being added, and then the square taken of the sum!!


Interference of de Broglie waves II
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Moreover, the wavelength of these waves in case of electrons is the de Broglie

wavelength =h/p

So can be understood electron interference pattern as arising from the interference of

de Broglie waves, each emanating from slit 1 or slit 2, i.e.

12(x, t) = 1(x, t) + 2(x, t)

and hence

P12(x) =|1(x, t) + 2(x, t)|2 = |1(x, t) + 2(x, t)|2

= |1(x, t)|2 + |2(x, t)|2 + 2Re

1(x, t)2(x, t)

=P1(x) + P2(x) + 2

P1(x)P2(x)cos .

Probabilities are time independent as we have averaged over a huge number ofelectron arrivals.

Note that we have taken the modulus squared since (with the benefit of hindsight) the

wave function often turns out to be complex.


Born interpretation of the wave function


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But 12(x, t)is the wave function at the point x at time t.

AndP12(x) =|12(x, t)|2 is the probability of the particle arriving at x (at timet)

So squaring the wave function a wave amplitude gives the probability of the particlebeing observed at a particular point in space.

This, and other experimental evidence, leads to the interpretation:

P(x, t) =|(x, t)|2dx =probability of finding the particle in region (x,x + dx)at timet.

This is the famous Born probability interpretation of the wave function.

AsP(x, t)is a probability, then (x, t) is usually called a probability amplitude (as well as thewave function).

From this interpretation follows the result that energy is quantized!


Probability interpretation of the wave function I
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40/94

The Born interpretation of the wave function (x, t)is

P(x, t) =|(x, t)|2

dx =probability of finding the particle in region (x,x + dx)at timet.

So what does probability mean?

To illustrate the idea, consider the followingsetup:

(

)ve charge

(+)ve mirror charge

x

liquid

helium

A negatively charged electron floats above a

sea of liquid helium.

It is attracted by its mirror image charge

But repelled by the potential barrier at the surfaceof the liquid

Classical physics says that it should come to rest

on the surface.

Quantum mechanics says otherwise.

Suppose we haveN =300 identical copies of thesame arrangement, and we measure the heightof the electron above the surface.


Probability interpretation of the wave function II
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Suppose we divide the distance above the surface of the liquid into small segments of

sizex

Measure the height of the electron in each of the

Nidentical copies of the experimentalarrangement.

LetN(x)be the number of electrons found to bein the range x to x + x, and form the ratio

N(x)

Nx

x

a0

no. of detections N(x) in in-terval(x,x + x)

0 28

0.5 69

1.0 76

1.5 582.0 32

2.5 17

3.0 11

3.5 6

4.0 2

4.5 1

Distance x from surface in units of a0

P(x, t) = |

(x, t)|

2

= 4(x

2

/a

3

0)e

2x/a0

0 0.5 1.0 1.5 .0 2.5 3.0 3.5 4.0 4.5 5.0

0

0.

1

0.2

0.3

0.4

0.

5

0.6N(x)

Nx

Data on the left plotted as a histogram.

The ratio

N(x)

N is the fraction of all the electronsfound in the range x to x + x.

In other words

N(x)

N probability of finding an electron in

the rangex to x + x


Probability interpretation of the wave function III
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So theprobabilityof finding the electron at a certain distance from the liquid helium

surface

is just the fractionof the times that it is found to be at that distance if we repeat themeasurement many many times over under identical conditions.

The results of the measurements will vary randomly from one measurement to the next.

According to the Born interpretation:

N(x)

Nprobability of finding an electron in the range x to x + xP(x, t)x |(x, t)|2x

where (x, t)is the wave function for the electron.

Since the electronmustbe found somewhere betweenand, thenall x

N(x)

N=

N

N=1


Probability interpretation of the wave function IV
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Or, in terms of the wave function, for any particle confined to the x axis, we must have:

+

|(x, t)

|2dx =1.

This is known as thenormalizationcondition.

All acceptable wave functions must be normalized to unity.

This means that the integral must converge

Which tells us that the wave functionmustvanish asx .

This is an important conclusion with enormous consequences.

There is one exception: the harmonic wave functions likeA cos(kx

t)which go on forever.

See later for their interpretation.

The requirement that (x, t)0 as x is known as aboundary condition.

Used when finding the solution to the Schrodinger equation.


Expectation value
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The results of measuring the position of a particle varies in a random way from one

measurement to the next.

We can use the standard tools of statistics to work out such things as the averagevalue of all the results, or their standard deviation and so on.

Thus, suppose we have data for the position of a particle in the form

N(x)

N= fraction of all measurements that lie in the range(x,x + x).

The average of all the results will then be

x allx

xN(x)

Nallx

xP(x, t)x

allxx|(x, t)|2x.

So, in the usual way of taking a limit to form an integral:

x =

+

x|(x, t)|2dx

known as the expectation value of the position of the particle.


Uncertainty
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We can also calculate the expectation value of any function of x:

f(x) = +

f(x)|(x, t)|2dx.

A most important example isx2:

x2

= +

x2

|(x, t)|2

dx

because then we can define theuncertaintyin the position of the particle:

(x)2 =(x x)2 =x2 x2.

This is a measure of how widely spread the results are around the average, orexpectation, valuex.

The uncertainty xis just the standard deviationof a set of randomly scattered results.


Example of expectation value and uncertainty I
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Can use the data from before (where a07.6nm)

x

a0

no. of detections N(x) in

interval(x,x + x)

0 28

0.5 69

1.0 76

1.5 58

2.0 32

2.5 17

3.0 11

3.5 6

4.0 24.5 1

The average value of the distance of the electron

from the surface of the liquid helium will be

x 0 28300

+ 0.5a0 69

300 + a0

76

300 + 1.5a0

58

30

+ 2a0 32

300 + 2.5a0

17

300 + 3a0

11

300 + 3.5a0

+ 4a0 2300

+ 4.5a0 1300

=1.235a0.

This can be compared with the result that follows for

the expectation value calculated from the wave

function for the particle:

x =

+

x |(x, t)|2 dx = 4

a30

0

x3 e2x/a0 dx =1.5a0


Example of expectation value and uncertainty II
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Similarly for the uncertainty: (x)2 allx

(x x)2 N(x)N

Using the data from earlier:

(x)2 (0 1.235)2a20 28

300 + (0.5 1.235)2a20

69

300

+ (1 1.235)2a20 76

300 + (1.5 1.235)2a20

58

300

+ (2 1.235)2a20 32300 + (2.5 1.235)2a20 17300

+ (3 1.235)2a20 11

300 + (3.5 1.235)2a20

6

300

+ (4 1.235)2a20 2

300 + (4.5 1.235)2a20

1

300

=0.751a20

which gives x =0.866a0.

The exact result using the wave function turns out to be the same.


Particle in infinite potential well I
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Consider a single particle of massm confined to within a region 0


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The first point to note is that, because of the infinitely high barriers, the particle cannot

be found in the regions x > L and x


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Between the barriers, the energy Eof the particle is purely kinetic, i.e. E =p2

2m.

Using the de Broglie relation p = kwe then haveE =

2k2

2m k =

2mE

We will usekto be the positive value i.e. k =

2mE

.

The negative value will be writtenk.

From the Einstein-Planck relationE = we also have =E/.

So we have the wave number(s) and the frequency of the wave function, but we have

to find a wave function that

Satisfies the boundary conditions (0, t) = (L, t) =0

And is normalized to unity.


Constructing the wave function II
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In the region0


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First, consider the boundary condition at x =0. Here, we must have

(0, t) =Aeit +Beit + Ceit +Deit

=(A +D)eit + (B + C)eit

=0.

This must hold true for all time, which can only be the case ifA +D =0and B + C =0.

Thus we conclude that we must have

(x, t) =Aei(kxt) +Bei(kxt) Bei(kx+t) Aei(kx+t)

=A(eikx eikx)eit B(eikx eikx)eit=2i sin(kx)(Aeit Beit).


Applying the boundary conditions IIQuantization of energy


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Now check on the other boundary condition, i.e. that (L, t) =0, which leads to:

sin(kL) =0

and hence

kL =n nan integer

This implies thatkcan have only a restricted set of values given by

kn = n

Ln =1, 2, 3, . . .

The negativen do not give a new solution, so we exclude them.

An immediate consequence of this is that the energy of the particle is limited to thevalues

En =

2k2n

2m=

2n22

2mL2 = n n =1, 2, 3, . . .

i.e. the energy is quantized.


Normalizing the wave function I


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Now check for normalization:

+ |(x, t)|

2

dx =4Aeit Beit2 L

0sin

2

(knx) dx =1

We note that the limits on the integral are(0,L)since the wave function is zero outside thatrange.

This integral must be equal to unity for all time, i.e. the integral cannot be a function of

time. But Aeit Beit2 = Aeit BeitAeit Beit=AA +BBABe2it ABe2it

which is time dependent!!

Can make sure the result is time independent by supposing either A =0or B =0.

It turns out that either choice can be made we will make the conventional choice and putB =0.



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Normalizing the wave function III

W h i l ill h /2 d h


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We can choose to suit ourselves so we will choose =/2and hence

A =

i 12L .The wave function therefore becomes

n(x, t) =

2

Lsin(nx/L)eint 0 < x


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The particle can only havediscreteenergiesEn

The lowest energy,E1 is greater than zero, as is required by the uncertainty principle.

Recall uncertainty in position is x 12Land uncertainty in momentum is pp, so, by the

uncertainty relation

xp p 2L

E 22

mL2E1.

Classically the particle can have any energy

0.

This phenomenon of energy quantization is to be found in all systems in which a

particle is confined by an attractive potential.

E.g. the Coulomb potential binding an electron to a proton in the hydrogen atom, or theattractive potential of a simple harmonic oscillator.

In all cases, the boundary condition that the wave function vanish at infinityguarantees

that only a discrete set of wave functions are possible.

Each allowed wave function is associated with a certain energy hence the energy

levels of the hydrogen atom, for instance.


Some Properties of Infinite Well Wave Functions
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The above wave functions can be written in the form

n(x, t) =

2

Lsin(nx/L)eiEnt/ =n(x)e

iEnt/

the time dependence is a complex exponential of the form exp[iEnt/]. The timedependence of the wave function for any system in a state of given energy is always of this

form.The factorn(x)contains all the spatial dependence of the wave function.

n(x) =

2

Lsin(nx/L)

Note a pairing of the wave functionn(x)with the allowed energy En. The wave function

n(x)is known as an energy eigenfunction and the associated energy is known as theenergy eigenvalue.


Probability Distributions I

Th b bilit di t ib ti di t th f ti bt i d b


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The probability distributions corresponding to the wave functions obtained above are

Pn(x) =|n(x, t)|2

=

2

Lsin(nx/L)eiEnt/

2

=2

Lsin2(nx/L) 0


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1(x) =

2

Lsin(x/L) P1(x) =

2

Lsin2(x/L)

0 L

x 0 L

x

Note that the probability is not uniform across the well. In this case there is a maximum

probability of finding the particle in the middle of the well.


Probability Distributions III

(x) =

2

sin(2x/L) P (x) =2

sin2(2x/L)
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2(x) =

L

sin(2x/L) P2(x) =L

sin (2x/L)

0 L

x

0 L

x

Once again, the probability distribution is not uniform.

However, here, there is are two maxima, atx =L/3and x =2L/3with a zero in

between.

Seems to suggest that the particle, as it bounces from side to side does not pass

through the middle!

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Probability Distributions V


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20(x) =

2

Lsin(20x/L) P20(x) =

2

Lsin2(20x/L)

0 L

x

0 L

x

Ifn becomes very large the probability oscillates very rapidly, averaging out to be 1/L,

so that the particle is equally likely to be found anywhere in the well.

This is what would be found classically if the particle were simply bouncing back and

forth between the walls of the well and observations were made at random times, i.e.

the chances of finding the particle in a region of sizexwillbex/L.


Expectation values and uncertainties I

Recall that if we were to measure the position of the particle in the well many many
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Recall that if we were to measure the position of the particle in the well many many

times

Making sure before each measurement that the particle has the samewave function (x, t)each time it is in the same state each time.

We get a random scatter of results whose average value will be

x =

+

x|(x, t)|2 dx

And whose standard deviation xis

(x)2 =(x x)2 =x2 x2

For a particle in an infinite potential well in the state given by the wave function n(x, t),

we have

|n(x, t)|2 = 2L

sin2(nx/L) 0


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I.e. the expectation value is in the middle of the well.

Note that this does not necessarily correspond to where the probability is a maximum. Infact, for, sayn =2, the particle is most likely to be found in the vicinity of x =L/4and

x =3L/4.

To calculate the uncertainty in the position, we need

x2 = 2L

L0

x2 sin2(nx/L) dx =L22n22 3

6n22 .

Consequently, the uncertainty in position is

(x)2 =x2 x2 =L2 n2

2

3n22

L2

4 =L2 n

2

2

612n22

.


Further developments
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The basic ideas presented till now can be extended in a number of ways:

Look at the properties of combinations of infinite square well wave functions, e.g.

(x, t) =12

(1(x, t) + 2(x, t))

This we can do already: find that this is not a stationary state, i.e. the probability|(x, t)|2

oscillates in time.

Work out expectation values of other quantities e.g. energy or momentum.

Work out the wave function for particles in other potentials, or for more complex

particles (e.g. atoms, molecules, solid state, gases, liquids, . . . )

The last set of aims can only be realised once we have the equation that the wave functionsatisfies: the Schrodinger wave equation.


Derivation of the Schrodinger wave equation I
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There is no true derivation of this equation.

But its form can be motivated by physical and mathematical arguments at a wide variety oflevels of sophistication.

Here we will derive the Schrodinger equation based on what we have learned so far

about the wave function.

The starting point is to work out, once and for all, what the wave function is for a

particle of energyEand momentump.

In the absence of any further information, we have used cos(kx t)and other harmonicfunctions.

But the correct result we get from the solution for the wave function of a particle in an infinitepotential well.


Derivation of the Schrodinger wave equation II

For the particle of energy E = and momentump = k(we wont worry about the n
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p gy p ( y

subscript here), this wave function is:

(x, t) =2

Lsin kxeit =i 1

2L

eikx eikx eit =i 1

2L

ei(kxt) ei(kx+t)

Classically, we would expect the particle to bounce back and forth within the well.

Here we have (complex) waves travelling to the left: ei(kx+t), and right: ei(kxt).

This suggests that the correct wave function for a particle of momentum p and energyEtravelling in the positivex direction is

(x, t) =Aei(kxt)

.

We want to find out what equation this particular function is the solution of.

All dynamical laws in nature involve rates of change e.g. velocity and acceleration inNewtons laws.

The rates of change here will be with respect to both time tand positionx, i.e. we will end upwitha partial differential equation.


Derivation of the Schrodinger wave equation III
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Taking derivatives with respect to position first gives:

2

x2 =A

2

x2ei(kxt) =k2Aei(kxt) =k2

This can be written, usingE =p2/2m = 2k2/2mas:

2

2m

2

x2

= p2

2m

.

Similarly, taking the derivatives with respect to time:

t=i

This can be written, usingE = as

i

t= =E.


Derivation of the Schrodinger wave equation IV

We now generalize this to the situation in which there is both a kinetic energy and a
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potential energy present, then E =p2/2m + V(x)so that

E = p2

2m + V(x)

where is now the wave function of a particle in the presence of a potential V(x).

Assuming that 2

2m

2

x2 =

p2

2mand i

t= =Estill hold true then we get

p2

2m + V(x) = E

2

2m

2

x2 + V(x) = i

t

Thus we arrive at 2

2m

2

x2 + V(x) =i

twhich is the famous time dependent

Schrodinger wave equation.


Some comments on the Schrodinger equation

The solution to the Schrodinger equation describes how the wave function changes as

f ti f d ti
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a function of space and time.

But it does not tell us how a particle moves through space.

Newtons equations tell us the position and velocity of a particle as it moves through space

Schrodingers equation tells us howthe information about the particlechanges with time.

Though called a wave equation, it does not always have wave-like solutions.

If the potential is attractive, i.e. tends to pull particles together, such as a Coulomb forcepulling an electron towards an atomic nucleus, then the wave function is not a wave at all:these are calledbound states.

In other cases, the particle is free to move anywhere in space: these are called unbound orscatteringstates.

There is a whole industry built around solving the Schrodinger equation in principle

all of chemistry flows from solving the Schrodinger equation.

We will concentrate on one part of solving the Schrodinger equation: obtaining the

wave function for a particle of a given energy E.


The Time Independent Schrodinger Equation I
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We saw that for the particle in an infinitely deep potential well that for a particle of

energyE, the solution looked like

(x, t) =

2

Lsin(kx)eiEt/ =(x)eiEt/

where the space and time parts of the wave function occur as separate factors.

This suggests trying a solution like this for the Schrodinger wave equation, i.e. putting

(x, t) =(x)eiEt/

If we substitute this trial solution into the Schrodinger wave equation, and make use of

the meaning of partial derivatives, we get:

2

2m

d2(x)

dx2 eiEt/ + V(x)(x)eiEt/ =i. iE/eiEt/(x) =E(x)eiEt/.


The Time Independent Schrodinger Equation II

The factorexp[iEt/]cancels from both sides of the equation, giving
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2

2m

d2(x)

dx2

+ V(x)(x) =E(x)

After rearranging terms:

2

2m

d2(x)

dx2 +

E V(x)(x) =0

which is thetime independent Schr odinger equation.

Note that the quantity E the energy of the particle is a free parameter in this

equation.

I.e. we can freely choose a value for Eand solve the equation.

Can emphasize this by writing the solution as E(x).

It appears that we would get a solution for any energy we choose, but it is not as

simple as that . . .


The nature of the solutions

The solutions to the Schrodinger equation assume two different forms:
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Those for bound states, i.e. where the particle is trapped by an attractive potential, e.g. theelectron in a hydrogen atom is bound to the proton by an attractive Coulomb potential.

The wave function must vanish as x .

This leads to the quantization of energy.

Those for scattering states where the particle is free to move through space, though itsbehaviour is influenced by the presence of some potential, e.g. an alpha particle scatteringoff a gold nucleus as in the original experiments by Rutherford.

The wave function does not vanish as x : in fact it looks something likeei(kxt).

Have to be careful (creative?) with the probability interpretation of .

In addition, there is a technical requirement: (x)and (x)must be continuous.

This is to guarantee that probability and the flow of probability behave as they should.

We will illustrate the two kinds of solutions in two important examples:

Harmonic oscillator

Scattering by a potential barrier.


The harmonic oscillatorSchrodinger equation

The quantum harmonic oscillator is one of the most important solved problems in
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q p p

quantum mechanics. It is the basis of the description of

Molecular vibrations.

Sound propagation in solids (phonons).

The quantum theory of the electromagnetic field.

The theory of the particles known as bosons.

Aclassicalharmonic oscillator consists of a particle of mass m subject to a restoring

force proportional to the displacement of the particle, i.e. F =kx.It undergoes simple harmonic motionx =A cos(t+ )of frequency =

k/m.

To determine the properties of a quantum simple harmonic oscillator, we need the

potential energy of such an oscillator. This is just V(x) = 1

2 kx2

=

1

2 m2

x2

.

The quantum mechanical simple harmonic oscillator of energyEis then described by

the Schrodinger equation

2

2m

d2

dx2 +

12

m2x2 =E


Allowed energies of the quantum harmonic oscillator

Recall thatEis a free parameter that we can choose as we like.
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Unfortunately, for almost all choices of Ewe run into an awkward problem: the solution

looks like

(x) e 12 (x/a)2 for x a =

m,

i.e. it diverges.

We shall see an example shortly.

So, whats wrong with that?

Recall that the solution must be normalized to unity, i.e. we must have +

|(x)|2 dx =1

and that can only happen if (x)

0 as x

.

Only if the energy is chosen to be

En =(n + 12

) n =0, 1, 2, . . .

do we get solutions that vanish asx

.


Unacceptable solutions of SHO Schrodinger equation

We can illustrate this by plotting the solutions as a function of E, starting atEless than5 d l l i i il h 5 (i )
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52and slowly increasing untilEreaches 5

2(i.e.n =2).

The solutions initially diverge tofor x but asE approaches 52, the solution

tends to extend along the x axis, ultimately asymptoting to zero along the x axis.

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The wave functions of a quantum SHO


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The wave functions for the allowed values of energy look like:

n(x) =

1

2nn!a

Hn(x/a)e

12

(x/a)2

whereHn(z)are known as the Hermite polynomials:

H0(z) =1 H1(z) =z H2(z) =z2

1 etc.

Forn =0, the solution is:

E0 = 1

2 0(x) =

1

a

e 1

2(x/a)2

P0(x) =|0(x)|2 = 1

a

e(x/a)

2

.

We can use this solution to show non-classical behaviour of a quantum harmonic

oscillator.


Non-classical properties of quantum SHO

The ground state probability distribution P0(x)and the potentialV(x)are plotted on the

same figure:
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g

P0(x) =1

ae

1

2

(x/a)2

V(x) = 12m

2x2

aa

classically

allowed region

classicallyforbidden region

classically

forbidden region

A classical oscillator of energy12 will oscillate betweenaand +a.

But the wave function is

non-zero forx >aand x


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e ot e c ass o p ob e s t at ca a se a e so ca ed scatte g p ob e s

Scattering problems involve particles that are not bound by an attractive potential.

The wave function does not vanish at infinity so the usual probability interpretation is nolonger valid, but we can still make sense of the solutions.

Find that quantum mechanics allows particles to access regions in space that are

classically forbidden.

I.e. according to classical physics, the particle does not have enough energy to get to someregions in space, but quantum mechanically, this has a non-zero probability of occurring.

Have already seen this occurring with the simple harmonic oscillator.

The possibility of a particle being found in forbidden regions in space gives rise tobarrier penetration and quantum tunnelling.

Shall look at the first in detail, the latter in outline.


Scattering by a potential barrierThe classical version

Shall consider a particle of energy Eincident on a potential barrier V0 withE


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Consider the classical version of the problem:

V(x)

x

V = 0

V0

incoming

particle

part c e re ecte

offbarrier

c ass ca y

forbidden region

for E< V0

The diagram shows a barrier defined by

V(x) =0 x0

Forx 0its energy will beE = p2

2m+ V0.

ButE 0 asit does not have enough energy.

The particle must simply bounce off the barrier.


Schrodinger equation for potential barrier

Recall that the time independent Schrodinger equation is given by
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2

2m

d2(x)

dx2 +

V(x)(x) =

E(x)

for a particle of energyEmoving in the presence of a potential V(x).

Here, the potential has two parts: V =0 for x 0.

The Schrodinger equation will also have two parts (with

d2

dx2 =

):

2

2m + 0. =E x< 0

2

2m + V0 =E x> 0

We have to solve both these equations, and then join the solutions together so that(x)and (x)are continuous.

We do not get energy quantization, but we do get other strange results.

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Solution of Schrodinger equation for potential barrier II

The physical meaning of this solution is best found by putting back the time

dependence.


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Recall that the solution to the time dependent wave equation for a particle of energyE is

(x, t) =(x)eiEt/ =(x)eit E =

In this case we have

(x, t) = Aei(kxt) + Bei(kx+t)

= wave travelling to the right + reflected wave travelling to the left

V(x)

V = 0

0

c ass ca y

forbidden region

for E < V0

incoming

wave

wave re ecte

offbarrier


Solution of Schrodinger equation for potential barrier III

For the region x > 0, we have to solve

2m

(V E) 0
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2

(V0E) =0

We are interested in the case in which E0

This is the case in which the particle has insufficient energy to cross over into x > 0.

So we put = 2m2

(V0

E) where >0

to give 2 =0

which has the solution

(x) =Cex +Dex CandD unknown constants.

We can check that this is indeed a solution by substituting the expression into the

equation for(x).


Solution of Schrodinger equation for potential barrier IV

There is a problem with this solution:
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As >0, the contributionex will grow to infinity asx

.

This would mean that the probability of finding the particle atx =would be infinite, whichis not physically acceptable.

This is a boundary condition once again.

So we must remove the ex contribution by puttingD =0 leaving:

(x) =Cex x> 0.

If we put the time dependence back in, we get

(x, t) =Cexeit

which is NOT a travelling wave.

The particle is not moving in the regionx > 0.


Matching the two solutions

Our overall solution so far is

(x) =Aeikx +Beikx x


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(x) =Cex x>0

These solutions must join together smoothly at x =0.

This requires(x)and (x)to be continuous at x =0.

The first matching condition gives

Aeikx +Beikxx=0 = Cexx=0i.e. A +B =C

The second matching condition gives

Aikeikx Bikeikxx=0 = Ce

x

x=0i.e. ik(A B) = CThese two equations can be easily solved for B and Cin terms ofA:

B = ik+

ik A C = 2ik

ik A

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Interpreting the solutions II

If these were water waves, or sound waves, or light waves, this would mean that the

waves are perfectly reflected at the barrier.

F b bilit lit d thi t th t th ti l h 100%


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For our probability amplitude waves, this seems to mean that the particle has a 100%chance of bouncing off the barrier and heading back the other way.

Recall: the particle has an energy which isless thanV0, so we would expect that theparticle would bounce off the barrier.

So this result is no surprise.

What we are actually doing here is calculating the particle flux associated with the incident

and reflected waves.

But the wave function is not zeroinside the barrier!

Forx >0we have

|(x)|2 =|A|2 2ikik ex2

=|A|2 4k2

2 + k2e2x.

This means that there is a non-zero probability of the particle being observed inside

the barrier. This is the region which is classically forbidden.

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An example I

Suppose a particle of energy Eapproaches a potential barrier of heightV0 =2E.


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The energy of the particle when it approaches the barrier is purely kinetic:

E =p2

2m p = 2mE

Using the de Broglie relation p = k, we get as beforek =

2mE

.

For thex >0 part of the wave function we have (using V0 =2E):

=

2m(V0E)

=

2mE

=k

Hence1

=k

1=

2 . The penetration depth is therefore

1

2a=

4 .

Thus, in this case of E = 12

V0, the penetration depth is roughly the de Broglie

wavelength of the incident waves. This gives a convenient estimate of the penetration

depth in different cases.


An example II

Consider two extreme cases:
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The particle is an electron moving at 1 ms1, then the penetration depth turns out to be

1

2 =0.06mm for an electron

The particle is a 1 kgm mass also moving at 1ms1. The penetration depth is then

1

2 =5.4 1032 m for a 1 kg mass.

Clearly, the macroscopic object effectively shows no penetration into the barrier

Hence barrier penetration is unlikely to ever be observed in the macroscopic world.

The microscopic particle penetrates a macroscopically measurable distance, and so

there is likely to be observable consequences of this.

Perhaps the most important isquantum tunnelling.


Quantum tunnelling
http://find/


94/94

If the barrier has only a finite width, then it is possible for the particle to emerge fromthe other side of the barrier:

V(x)

xV = 0

V0

incoming

wave

wave reflected

offbarrier

transmitted

wave

http://find/

wave mechanics lecture

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